What is the derivative of 4x(x^2-1) ?

The derivative of 4x(x^2-1) is 4(3x^2-1)

What is the first derivative of 4x(x^2-1) ?

The first derivative of 4x(x^2-1) is 4(3x^2-1)

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derivative of

4 x (x^{2} - 1)

4 (3 x^{2} - 1)

Solution steps

\frac{d}{d x} (4 x (x^{2} - 1))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 4 \frac{d}{d x} (x (x^{2} - 1))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = x^{2} - 1

= 4 (\frac{d x}{d x} (x^{2} - 1) + \frac{d}{d x} (x^{2} - 1) x)

\frac{d x}{d x} = 1

\frac{d}{d x} (x^{2} - 1) = 2 x

= 4 (1 \cdot (x^{2} - 1) + 2 xx)

Simplify

4 (1 \cdot (x^{2} - 1) + 2 xx) : 4 (3 x^{2} - 1)

= 4 (3 x^{2} - 1)

\frac{d y}{d t} = 16 - \frac{y}{250 + 0 t}

y = 3 x^{2} + 10 x - 26

\int sin^{5} (x) cos (x) d x

\frac{d}{d x} (x^{s i n (2 x)})

\frac{d}{d x} (sin^{2} (x) - cos^{2} (x) sin^{2} (x))

What is the derivative of 4x(x^2-1) ?
The derivative of 4x(x^2-1) is 4(3x^2-1)
What is the first derivative of 4x(x^2-1) ?
The first derivative of 4x(x^2-1) is 4(3x^2-1)